Geometry
Find the Area of triangle, 2 and 5 are the radii of the semicircles.
1. 625/6
2. 625/3√10
3. 576
4. 576/3√10
See our previous ‘Questions of the Day’:
Quant Question Of The Day: 156
Quant Question Of The Day: 155
Question of the day
Counting
A group of students were given the names of eight countries and their current presidents in two columns. Each student is expected to match the eight countries with their respective presidents. A student is awarded 3 marks for each correct match. What is the maximum possible number of students who give different sets of valid responses and end up getting 15 marks?
1. 56
2. 112
3. 56×3!
4. 56×3!/2!
Out of 8 select 5 correctly matched in 8C5 ways.
Each of the remaining three countries are incorrectly matched.
this can happen in 2 ways.
Dearrangement of 3 = 2
The total number of ways in which students can get 15 marks is 8C5 x 2= 112 ways.
See our previous ‘Questions of the Day’:
We have two closed boxes with a label on each one. Label A on box A: “The label on box B is true and the gold is in box A” Label B on box B: “The label on box A is false and the gold is in box A” We don’t know if the labels (one or both of them) are true or false. Knowing FOR SURE that the gold is in box A or in box B, where’s the gold?
Algebra
How many real numbers x are solutions to the following equation?
2017x + 2018x = 2019x .
Jeevan Ram a farmer has some strange animals. His hens have 2 heads and 8 legs, his peacocks have 3 heads and 9 legs, and his zombie hens have 6 heads and 12 legs. Jeevan Ram counts 800 heads and 2018 legs on his farm. What is the number of animals that Jeevan Ram has on his farm?
How many triplets of non-negative integers (x,y,z) satisfying the equation xyz + xy + yz + zx + x + y + z = 2014
1. 18
2. 24
3. 27
4. 36
5. None of the above
xyz + xy + yz + zx + x + y + z = 2014 xyz + xy + yz + zx + x + y + z + 1 = 2015 ( x+1) ( y+1) ( z+1) = 2015 = 5 × 13 × 31 If all (x, y , z ) are positive then there are 3! = 6 solution If exactly one of x, y and z is 0, there are 3 18 solutions. If exactly two of x, y and z zero then there are 3 solutions. Total: – 6 + 18 + 3 =27 solutions.
There are five carton box of different weight. They are weighed in pairs of two with all possibilities. The weights in kgs are 165, 168, 169.5, 171, 172.5, 174, 175.5, 177, 180, and 181.5 . How much does each container weigh?
Lets assume that the weights of five containers are B1, B2, B3, B4 and B5 kg respectively. Also, B1 ≤ B2 ≤ B3 ≤ B4 ≤ B5. It is given that five cartons of oil are weighed two at a time in all possible ways. It means that each of the container is weighted four times. Thus, 4×(B1 + B2 + B3 + B4 + B5) = (165 + 168 + 169.5 + 171 + 172.5 + 174 + 175.5 + 177 + 180 + 181.5) 4×(B1 + B2 + B3 + B4 + B5) = 1734 (B1 + B2 + B3 + B4 + B5) = 433.5 kg……….(1) Now, B1 and B2 must add to 165 as they are the lightest one. B1 + B2 = 165……………(2) Similarly, B4 and B5 must add to 181.5 as they are the heaviest one. B4 + B5 = 181.5………….(3) From above three equation, we get B3 = 87 kg Also, it is obvious that B1 and B3 will add to 168 – the next possible higher value. Similarly, B3 and B5 will add to 180 – the next possible lower value. B1 + B3 = 168………….(4) B3 + B5 = 180………….(5) Substituting B3 = 87, we get B1 = 81 and B5 = 93 From 2 & 3 equations, we get B2 = 84 and B4 = 88.5 Hence, the weights of five containers are 81, 84, 87, 88.5 and 93 kg.
See our previous ‘Puzzles of the Day’:
Numbers
Find the sum of all four-digit numbers N whose sum of digits is equal to 2010 – N.
N has to be greater than 2010 – 28 = 1982 because maximum sum of digits less than 2010 is 1 + 9 + 9 + 9 i.e. 28. Now checking for all the decades there are only two possible values of N i.e. 1986 and 2004. So the required sum is 3990.
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